Discussion Forum

let $f(x)=(1-x)^{2}\sin^{2} x+x^{2}$ for all x $\epsilon R$ and 

let $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-ln t\right) f(t) dt,$ for all x $\epsilon (1,\infty)$ 

which of the following is true?

Answer:

Explanation:

 Concept involved use of Newton leibnits formula

 i.e, $\frac{d}{dx}\left\{\int_{g(x)}^{f(x)} \phi(t) dt\right\}=\phi (f(x)).f'(x)-\phi (g(x)).g'(x)$

  Sol. Here , $f(x)(1-x)^{2} \sin^{2}x+x^{2},$ for all x

     $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log(t)\right)f(t)dt$

$\forall x \epsilon (1, \infty)$

Here, 

$f(x)=(1-x)^{2}\sin^{2} x+x^{2}$ for all x 

 and  $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log t \right) f(t) dt,$ 

 $\Rightarrow$   $g'(x)=\left\{\frac{2(x-1)}{x+1}-\log x\right\}f(x) (+ve)....(i)$

 for   g'(x) to be increasing or decreasing 

Let $\phi(x)=\frac{2(x-1)}{(x+1)}-\log x$

 $\phi'(x)= \frac{4}{(x+1)^{2}}-\frac{1}{x}=\frac{-(x-1)^{2}}{x(x+1)^{2}}$

 $\phi '(x) <0,$ for x >1

 $\Rightarrow$  $\phi (x) < \phi(1) \Rightarrow \phi (x) <0$......(ii)

 For  Eqs.(i) and (ii) , we get

 $g'(x) <0$ for $x \epsilon (1 ,\infty)$

 $\therefore$ g(x) is decreasing for $x \epsilon (1 ,\infty)$